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\title{离散数学作业(习题四)}
\author{2020141460280张家帅}

\begin{document}
\maketitle
\section*{习题四}
\subsection*{T(16)}
\subsubsection*{(1)}
\begin{align*}
	\mbox{左边} & = r(R_1\cup R_2)              \\
	            & = R_1\cup R_2\cup I_A         \\
	            & = R_1\cup I_A\cup R_2\cup I_A \\
	            & = r(R_1)\cup r(R_2)           \\
	            & = \mbox{右边}
\end{align*}
\begin{align*}
	\mbox{左边} & = r(R_1\cap R_2)\notag                 \\
	            & = (R_1\cap R_2)\cup I_A\notag          \\
	            & = (R_1\cup I_A)\cap(R_2\cup I_A)\notag \\
	            & = r(R_1)\cap r(R_2)\notag              \\
	            & =\mbox{右边}
\end{align*}
\subsubsection*{(3)}
由定义得：
\[t(R_1\cup R_2)=\bigcup\limits^n_{i=1}(R_1\cup R_2)^i\]
\[t(R_1)\cup t(R_2)=\bigcup\limits^n_{i=1}R_1^i\cup\bigcup\limits^n_{i=1}R_2^i\]

下面利用数学归纳法证明：$\bigcup\limits^n_{i=1}(R_1\cup R_2)^i\supseteq  \bigcup\limits^n_{i=1}R_1^i\cup\bigcup\limits^n_{i=1}R_2^i$

\ding{192} $n=1$时： $R_1\cup R_2\supseteq R_1\cup R_2$成立

\ding{193} $n=k$时：假设$(R_1\cup R_2)^k\supseteq R_1^k\cup R_2^k$成立，考虑$n=k+1$时的情况：
\begin{align*}
	(R_1\cup R_2)^{k+1} & =(R_1\cup R_2)\circ(R_1\cup R_2)^k                                       \\
	                    & \supseteq (R_1\cup R_2)\circ(R_1^k\cup R_2^k)                            \\
	                    & =R_1\circ R_1^k\cup R_1\circ R_2^k\cup R_2\circ R_1^k\cup R_2\circ R_2^k \\
	                    & =R_1^{k+1}\cup R_2^{k+1}\cup R_1\circ R_2^k\cup R_2\circ R_1^k           \\
	                    & \supseteq  R_1^{k+1}\cup R_2^{k+1}\mbox{（成立）}
\end{align*}
综上所述，$\bigcup\limits^n_{i=1}(R_1\cup R_2)^i\supseteq  \bigcup\limits^n_{i=1}R_1^i\cup\bigcup\limits^n_{i=1}R_2^i$成立

原式得证
\end{document}